def spynum (n,r,sm=0,pd=1):
#s = 0
#p = 1
#b = False
if n == 0:
print (sm==pd)
return sm == pd
else:
rem = n % 10
sm += rem
pd *= rem
print (sm,pd,rem)
spynum(n//10,rem,sm,pd)
num = 1124
print (spynum(num,num%10,0,1))
The program returns Boolean also working if I print the variable inside base condition but it is not print the same outside the function.Im really confused ABT it !
def spynum (n,r,sm=0,pd=1):
#s = 0
#p = 1
#b = False
if n == 0:
print (sm==pd)
return sm == pd <-- returns either true or false
else:
rem = n % 10
sm += rem
pd *= rem
print (sm,pd,rem)
spynum(n//10,rem,sm,pd)
num = 1124
print (spynum(num,num%10,0,1))
The == operator compares and returns either a true or false value. To return a number, consider returning an int value.
Recursive functions work exactly like non-recursive functions; they return to their immediate caller, and if you don't return a value on some path, they return None.
You need to return the recursive result:
def spynum (n, r, sm=0, pd=1):
if n == 0:
print (sm == pd)
return sm == pd
else:
rem = n % 10
sm += rem
pd *= rem
print (sm, pd, rem)
return spynum(n//10, rem, sm, pd)
spynumfrom insidespynumfrom insidespynum... and so on. At some point the last invocation finds that n == 0 and returns a boolean. That boolean is returned to the invocation before it, which simply ignores it, not to the main program. You need to propagate the answer back through your chain ofspynuminvocations, all the way to the top and the main program. It's probably just a one-word change: addreturnbefore the recursive invocation ofspynum.